The calculation basis of the heat output of components in the cabinet is as follows: (related to the installation of components)

1) Calorific value of frequency converter, transformer, driver, servo amplifier, etc.: calorific value of Rated Power 1KW is about 30 ~ 50W (divided into fan pump load and mechanical load depending on load condition);

2) The calorific value of PLC is about 35 ~ 50W (in groups); the calorific value of industrial computer is calculated according to the size of industrial computer. Generally about 300W / set;

3) Heating capacity of contactor and other components: the rated power is 1kW, about 5 ~ 20W. Compared with high-power heating components, it can be ignored.

4) Heating capacity of general server: about 280-500w; heating capacity of UPS: 20% of power

For example, when the frequency converter works at full load, its total loss (converted into heat) is about 3% ~ 5% of the rated power of the system. It can be calculated that the loss is about 30W ~ 50W when the 1kW frequency converter works at full load.

5) The heating of SCR is 2W / A; the DC driver is 1kW, about 7W ~ 10W.

The formula to be used is as follows:

Q t = (Q i + Q r )*1.2

Q T: total heat generated by cabinet (unit: W)

Q I: total heat generated in cabinet (unit: W)

Q R: heat transferred from the outside of the cabinet to the inside of the cabinet (unit: W)

Q I: total heat generated in cabinet (unit: W)

Q r = k * a * Δ t k: heat transfer coefficient

1) K = 5.5w/m2. K steel cabinet

2) K = 12.0w/m2. K aluminum magnesium alloy cabinet

3) K = 0.2w/m2. K plastic material cabinet

A: Surface area of cabinet (unit: m2)

Δ t = t 1 - T 2 (unit: ℃)

T 1: maximum temperature outside the cabinet T 2: control temperature inside the cabinet

For example, the external dimension of a steel cabinet is: length * height * thickness: 1500 * 2000 * 800 mm, the heat generated by the heating elements inside the cabinet is 1000W, the control temperature inside the cabinet is 28 ℃, and the outside temperature is 35 ℃. How much cooling capacity is suitable for this cabinet?

Solution: the surface area of the cabinet is a = 1.5 * 2 * 2 + 0.8 * 2 * 2 + 1.5 * 0.8 = 10.4 m2

Heat transferred from the outside of the cabinet to the inside of the cabinet: Q r = k * a * Δ t = 5.5 * 10.4 * (35-28) = 400.4w

Total heat generated by cabinet: Q T = (Q I + Q R) * 1.2 = (1000 + 400.4) * 1.2 = 1680.48 w

Therefore, the cabinet air conditioner with a refrigerating capacity of 2000W is selected.